For final value theorem:
Suppose $X(s)$ is rational and strictly proper, then if $X(s)$ has all poles in the open left half plane and at most one pole at zero, then $$\lim\limits_{t \to \infty} x(t) = \lim\limits_{s \to 0} s X(s)$$
I am trying to reverse engineer the proof and I couldn't figure out how the properties on the poles came into play.
Proof: $\mathcal{L}(\dot x(t)) = sX(s) - x(0) = \int\limits_0^\infty \dot x(t) e^{-st}\mathrm{d}t$, take limit on both sides as $s \to 0$, we have $\lim\limits_{s \to 0} s X(s) = \int\limits_0^\infty \dot x(t) \mathrm{d}t + x(0) = \lim\limits_{t \to \infty} x(t)$. End of proof.
It is not obvious to me how the conditions on the poles came into play here.
Suppose $X(s)$ had pole of order $\geq 2$ at zero i.e. $X(s) = \dfrac{1}{s^2}$, then it is clear that the left hand side will blow up thus not yielding a finite value. But this does not mean that $\lim\limits_{t \to \infty} x(t) \to \infty$.
Suppose $X(s)$ had a pole in the right half plane (i.e. real part of pole $> 0$), then it seems that the proof will still hold.
Can someone explain how one would obtain the conditions on the poles for the final value theorem?
Let's do it by parts.
First assume $ X(s) $ has two or more poles at zero, then $ \underset{s \rightarrow 0}{\lim} sX(s) \rightarrow \infty $. While $ \underset{t \rightarrow \infty}{\lim} x(t) $ can either tend to infinity or to an arbitrary value depending on the initial state.
Second assume $ X(s) $ has at least one pole with positive real part. Then $ \underset{s \rightarrow 0}{\lim} sX(s)$ tends to some constant while $\underset{t \rightarrow \infty}{\lim} x(t) \rightarrow \infty $.
You can verify this two conditions by taking the inverse Laplace transform of $ X(s) $, and verify the conditions for the resulting ODE to converge to $ \underset{s \rightarrow 0}{\lim} sX(s) $.
EDIT: As an addendum, this are the conditions where the equality you used hold.